details

property	value
status	complete
benchmark	213407.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n027.star.cs.uiowa.edu
space	ICFP_2010

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	9.70746898651 seconds
cpu usage	35.13470547
max memory	3.266002944E9

stage attributes

key	value
output-size	22148
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 682 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 1 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) DependencyGraphProof [EQUIVALENT, 0 ms] (9) QDP (10) MRRProof [EQUIVALENT, 236 ms] (11) QDP (12) PisEmptyProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPOrderProof [EQUIVALENT, 36 ms] (18) QDP (19) PisEmptyProof [EQUIVALENT, 0 ms] (20) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(1(1(x1))) -> 1(0(2(1(x1)))) 0(1(1(x1))) -> 1(1(0(3(2(x1))))) 0(1(1(x1))) -> 1(4(0(0(2(1(x1)))))) 0(1(4(x1))) -> 1(4(0(3(x1)))) 0(1(4(x1))) -> 4(0(2(1(x1)))) 0(1(4(x1))) -> 4(0(2(1(3(x1))))) 0(4(1(x1))) -> 0(2(4(1(x1)))) 0(4(1(x1))) -> 4(0(2(1(x1)))) 0(4(1(x1))) -> 4(0(3(1(x1)))) 0(4(1(x1))) -> 1(4(4(0(2(x1))))) 0(4(1(x1))) -> 2(1(4(0(2(x1))))) 0(5(4(x1))) -> 1(4(0(0(2(5(x1)))))) 0(5(4(x1))) -> 4(0(2(5(2(5(x1)))))) 0(5(4(x1))) -> 5(0(4(0(3(3(x1)))))) 4(2(1(x1))) -> 4(0(2(1(x1)))) 4(2(1(x1))) -> 1(2(4(0(2(x1))))) 0(1(0(4(x1)))) -> 0(0(2(2(1(4(x1)))))) 0(1(1(2(x1)))) -> 5(1(0(2(1(x1))))) 0(1(1(3(x1)))) -> 1(1(5(0(3(x1))))) 0(1(4(1(x1)))) -> 1(0(2(4(1(x1))))) 0(1(4(3(x1)))) -> 1(0(2(4(3(x1))))) 0(1(4(3(x1)))) -> 4(0(2(1(3(x1))))) 0(4(1(2(x1)))) -> 2(0(3(1(4(x1))))) 0(4(2(1(x1)))) -> 0(4(3(0(2(1(x1)))))) 0(5(0(4(x1)))) -> 1(5(4(0(0(2(x1)))))) 0(5(1(3(x1)))) -> 1(1(5(0(3(x1))))) 0(5(1(3(x1)))) -> 5(3(1(0(3(x1))))) 0(5(4(1(x1)))) -> 4(1(5(4(0(2(x1)))))) 0(5(4(3(x1)))) -> 0(2(5(0(3(4(x1)))))) 0(5(4(3(x1)))) -> 1(5(3(4(0(3(x1)))))) 1(0(5(4(x1)))) -> 1(4(5(0(3(3(x1)))))) 1(0(5(4(x1)))) -> 5(5(1(0(2(4(x1)))))) 1(4(2(1(x1)))) -> 4(0(2(1(2(1(x1)))))) 4(1(2(1(x1)))) -> 4(1(0(2(1(x1))))) 4(1(2(1(x1)))) -> 3(4(0(2(1(1(x1)))))) 4(3(2(1(x1)))) -> 0(3(4(0(2(1(x1)))))) 0(0(1(2(3(x1))))) -> 1(0(3(0(0(2(x1)))))) 0(0(5(1(2(x1))))) -> 0(0(2(5(0(1(x1)))))) 0(0(5(1(3(x1))))) -> 4(5(0(0(3(1(x1)))))) 0(1(3(4(2(x1))))) -> 1(3(4(3(0(2(x1)))))) 0(4(5(3(4(x1))))) -> 0(3(2(5(4(4(x1)))))) 0(5(0(2(2(x1))))) -> 0(0(2(5(2(4(x1)))))) 0(5(1(1(3(x1))))) -> 0(3(2(1(1(5(x1)))))) 0(5(1(4(3(x1))))) -> 3(0(2(1(5(4(x1)))))) 0(5(5(1(2(x1))))) -> 1(5(3(0(2(5(x1)))))) 1(0(1(2(4(x1))))) -> 1(1(0(2(3(4(x1)))))) 1(4(2(1(2(x1))))) -> 0(2(2(1(1(4(x1)))))) 4(0(0(5(4(x1))))) -> 5(0(0(4(4(5(x1)))))) 4(2(5(4(1(x1))))) -> 4(4(1(5(3(2(x1)))))) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- popout

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all output return to SRS Stand 10685